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Wolfram gives this nice result: $$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$

I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting very complicated. Any help/hints ?

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    $\begingroup$ Write $\cos 2x = 1 - 2\sin^2 x$, and use the substitution $\sin x = t$. Then $\cos x \, dx = dt$ $\endgroup$ – M. Vinay Sep 6 '14 at 13:38
  • $\begingroup$ I have tried that already, its not any better than weiestrass substitution :( $\endgroup$ – rrr Sep 6 '14 at 13:39
  • $\begingroup$ I get : $$\int \dfrac{1}{(1-2t^2)\sqrt{1-2t^2}} dt $$ then I am clueless on how to reach the wolfram's answer from here. By looking at the answer, I feel there should be some trick/simpler method $\endgroup$ – rrr Sep 6 '14 at 13:41
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    $\begingroup$ $\cos x = \cos (2x - x) = \cos 2x \cos x + \sin 2x \sin x$. $\endgroup$ – M. Vinay Sep 6 '14 at 13:44
  • $\begingroup$ Interesting, does that give : $$\int \dfrac{\cos x}{\sqrt{\cos 2x}} dx + \int \dfrac{\tan 2x \sin x}{\sqrt{\cos 2x}}dx$$ ? $\endgroup$ – rrr Sep 6 '14 at 13:46
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$\cos2x=1-2\sin^2 x$, let $t=\sin x$ so $dt=\cos xdx$ in numerator. SO: $$I=\int\frac{dt}{(1-2t^2)^{3/2}}$$ Then use the substitution $u^2=(1-2t^2)$ so $udu=-2tdt$, so $\displaystyle dt=-\frac{udu}{2t}=\frac{-udu}{\sqrt2\sqrt{1-u^2}}$: $$I=-\int\frac{udu}{\sqrt2\sqrt{1-u^2}u^3}=\frac{-1}{\sqrt2}\int\frac{du}{u^2\sqrt{1-u^2}}$$ Then use $v=1/u$ so $\displaystyle du=-\frac1{v^2}dv$ So: $$I=\frac1{\sqrt2}\int\frac{vdv}{\sqrt{v^2-1}}=\frac1{\sqrt2}\sqrt{v^2-1}+C$$ Now: $$\sqrt{v^2-1}=\frac{\sqrt{1-u^2}}u=\frac{\sqrt2t}{\sqrt{1-2t^2}}=\frac{\sqrt2\sin\theta}{\sqrt{1-2\sin^2\theta}}=\frac{\sqrt2\sin\theta}{\sqrt{\cos2\theta}}$$ So: $$\large I=\frac{\sin\theta}{\sqrt{\cos2\theta}}+C$$

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  • $\begingroup$ @rational answer edited $\endgroup$ – RE60K Sep 6 '14 at 13:47
  • $\begingroup$ Thank you so much :) I'm still going through it, this seems extension to my attempt $\endgroup$ – rrr Sep 6 '14 at 13:49
  • $\begingroup$ @rational refresh to see that I removed last substitution $\endgroup$ – RE60K Sep 6 '14 at 13:50
  • $\begingroup$ I see... looks perfect. thanks a lot !! $\endgroup$ – rrr Sep 6 '14 at 13:54
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Alternative method (akin to finding an integrating factor by inspection):

$\cos x = \cos (2x - x) = \cos 2x \cos x + \sin 2x \sin x$.

Then the integral becomes:

$$\begin{align} \int \dfrac{\cos 2x \cos x + \sin 2x \sin x}{\cos^{\frac 3 2} 2x} dx & = \int \dfrac{\cos x}{\sqrt {\cos 2x}} + \dfrac{\sin x \sin 2x}{\cos^\frac 3 2 2x}dx\\ & = \int \dfrac{d (\sin x)}{\sqrt {\cos 2x}} - \dfrac 1 2 \int \dfrac{\sin x \, d(\cos 2x)}{\cos^\frac 3 2 2x}\\ & = \int \dfrac{du}{\sqrt v} - \dfrac 1 2 \int\dfrac{u\, dv}{v^{3/2}} \qquad [u = \sin x,\ v = \cos x]\\ & = \int d\left( \dfrac{u}{\sqrt v} \right)\\ & = \dfrac{u}{\sqrt v}\\ & = \dfrac{\sin x}{\sqrt {\cos 2x}} \end{align}$$

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  • $\begingroup$ brilliant ! I like this method more xD but i have accepted the answer already and hate to undo sorry :( $\endgroup$ – rrr Sep 6 '14 at 14:05
  • $\begingroup$ Good, don't. I don't really like this answer. I had already typed half of it when the other answer came up, so I didn't discard it. $\endgroup$ – M. Vinay Sep 6 '14 at 14:06
  • $\begingroup$ Hey, isn't the second expression missing a $1/2$ factor? $d(cos2x)=-2sin2x$. $\endgroup$ – A Googler May 8 '15 at 17:45
  • $\begingroup$ @AGoogler Thanks, I corrected it! $\endgroup$ – M. Vinay May 9 '15 at 2:10
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$$\int\dfrac{\cos xdx}{\cos^{3/2}2x}=\int\dfrac{\cos xdx}{(\cos^2x-\sin^2x)^{3/2}}=$$ $$\dfrac{\cos xdx}{(1-\tan^2x)^{3/2}\cos^3x}=\int\dfrac{\sec^2xdx}{(1-\tan^2x)^{3/2}}$$ $$u=\tan x,du=\sec^2xdx$$ $$\int\frac{du}{(1-u^2)^{3/2}}$$ $$u=\sin\theta,du=\cos\theta d\theta$$ $$\int\dfrac{\cos\theta d\theta}{\cos^3\theta}=\tan\theta+C=\frac{u}{\sqrt{1-u^2}}+C=\frac{\tan x}{\sqrt{1-\tan^2x}}+C=$$ $$\dfrac{\sin x}{\sqrt{\cos^2x-\sin^2x}}+C=\dfrac{\sin x}{\sqrt{\cos2x}}+C$$

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  • $\begingroup$ I like this best. $\endgroup$ – M. Vinay Sep 6 '14 at 15:36

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